Question

Identify final products in the following sequence of reactions
\(\text{RCOOH}+\text{R’OH}\overset{\text{H}^+\,}{\rightarrow }\text{intermediate}\overset{\text{Ni},\Delta \,}{\rightarrow }^{\text{H}_2}\text{Products}\)

• A. \(\text{R’COOH}+\text{RCH}_2\text{OH}\)
• B. \(\text{RCH}_2\text{OH}+\text{R’}-\text{OH}\)
• C. \(\text{R’CH}_2\text{OH}+\text{R}-\text{OH}\)
• D. \(\text{R}-\text{OH}+\text{R’}-\text{OH}\)

Hint:

Hydrogenation cleaves an ester into two alcohols.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A carboxylic acid and an alcohol in the presence of \(\text{H}^+\) give an ester: \(\text{RCOOH} + \text{R}'\text{OH} \to \text{RCOOR}' + \text{H}_2\text{O}\). The ester is the intermediate.

Step 2: Key Formula or Approach:
Catalytic hydrogenation with \(\text{H}_2\) over Ni on heating reduces an ester to two alcohols: \(\text{RCOOR}' + 2\text{H}_2 \to \text{RCH}_2\text{OH} + \text{R}'\text{OH}\).

Step 3: Detailed Explanation:
The acyl part \(\text{RCO}-\) is reduced to the primary alcohol \(\text{RCH}_2\text{OH}\).
The alkoxy part \(\text{R}'\text{O}-\) is released as the alcohol \(\text{R}'\text{OH}\).
So the final products are \(\text{RCH}_2\text{OH}\) and \(\text{R}'\text{OH}\). Option A swaps the groups, and the acid in A is not made. Options C and D do not reduce the carbonyl.

Final Answer:
The products are \(\text{RCH}_2\text{OH}\) and \(\text{R}'-\text{OH}\), option (B). \[ \boxed{\text{RCH}_2\text{OH} + \text{R}'\text{OH}} \]